Problem Set #13 Solutions

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8.01L: Physics I January 13, 2016 Prof. Alan Guth

Problem Set #13 Solutions

Due by 11:00 am on Friday, January 22 in the bins at the intersection of Buildings 8 (3rd floor) and 16 (4th floor). Please put your name and recitation number clearly on the top page of your problem set.

SOURCES:

University Physics, Volume 1, 13th Edition, by Hugh D. Young and Roger A. Freed-man (Addison-Wesley, 2012). This is the required textbook for the course.

Essentials of Introductory Classical Mechanics, 6th Edition, by Wit Busza, Susan Cartwright, and Alan H. Guth, now available to EVERYBODY at http://web.mit.edu/ 8.01L/studyguide/index.html. Also known as the BCG Study Guide.

READING: Chapter 10, Sections 10.4–10.7 (DYNAMICS OF ROTATIONAL MOTION: Work and Power in Rotational Motion; Angular Momentum; Conservation of Angular Mo-mentum; Gyroscopes and Precession) and Chapter 13, Sections 13.1–13.8: (GRAVITATION: Newton’s Law of Gravitation; Weight; Gravitational Potential Energy; The Motion of Satel-lites; Kepler’s Laws and the Motion of Planets; Spherical Mass Distributions; Apparent Weight and the Earth’s Rotation; Black Holes).

OPTIONAL ADDITIONAL READING: Busza, Cartwright, and Guth (BCG), Chap-ter 9, pp. 311–322. (“Optional” means that you should read this only if you find it useful.) NOTE: Your written solutions must include a brief commentary in addition to any equations or graphs used to arrive at your answer. For example, this commentary should explain your basic strategy for solving the problem and also highlight the concept before applying an equation. We also want you to first solve a problem algebraically before you put in the particular numbers relevant to the problem. We’ll say that again: solve the answer in terms of variables before you start sticking in numbers! This makes it easier for you when you check your work and also for the grader to follow your logic.

13-1: Square pendulum

Young and Freedman (13th edition) problem 14.79, page 468. (same as Young and Freedman (12th edition) problem 13.71, page 452.) In doing this problem, you may want to refer to Section 14.6, The Physical Pendulum, pp. 455–456 of the 13th edition (Section 13.6, pp. 438–39 in 12th edition).

A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner. If it is rotated slightly to the left and then released, at what frequency will it swing back and forth?

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Solution:

This is a physical pendulum, the frequency of oscillation is given by f = 1

2π r

mgd

I (1)

The center of mass of the square object is simply its geometrical center, d = L cos 45◦ = √L

2. (2)

Now we need to calculate the moment of inertia of the square I about an axis through the pivot, perpendicular to the plane of the square. By parallel-axis theorem,

I = Icm+ md2 = Icm+

mL2

4 . (3)

To calculate Icm, think of the square as 4 sticks, each stick contributes a moment of

inertia of ₁₂1 m₄ L2₊ m 4

L 2

2

. Again by parallel-axis theorem,

I = 4 1 12 m 4 L2+ m 4 L 2 2! + mL 2 2 (4) = 5 6mL 2 (5) Substitute into (1) gives

f = 1 2π s mg√L 2 5 6mL2 = 1 2π s 6 5√2 r g L. (6)

13-2: Parachutist and turntable

Based on Young and Freedman (13th edition) exercise 10.45, page 337. (same as Young and Freedman (12th edition) exercise 10.43, page 347.)

A large wooden turntable in the shape of a flat uniform disk has a radius R and a total mass M . The turntable is initially rotating at angular frequency ω0 about a vertical

axis through its center. Suddenly, a parachutist of mass m makes a soft landing on the turntable at a point near the outer edge.

(a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

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Solution:

Since there is no net external torque around the rotation axis, angular momentum of the system of parachutist and turntable is conserved. With L0 = I0ω0 and

L1 = I1ω1, L0 = L1 implies

ω1 =

I0

I1

ω0 (7)

Since I0 = 1₂M R2 and I1 = I0+ mR2 we have

I0 I1 = 1 1 + 2m_M . (8) Therefore ω1 = 1 1 + 2×70.0 kg_{120 kg} × 3.00 rad/s = 1.38 rad/s. (9) (b) Compute the kinetic energy of the system before and after the parachutist lands.

Why are these kinetic energies not equal? Solution:

The kinetic energy of the system before the parachutist lands K0 = 1 2I0ω 2 0 = 1 2 1 2M R 2 ω2₀ (10) = 1 4× 120 kg × (2.00 m) 2_{× (3.00 rad/s)}2 ₍₁₁₎ = 1080 J (12)

after the landing, K1 = 1 2I1ω 2 1 = 1 2I1 I0 I1 ω0 2 = I0 I1 K0 = 1 1 + 2m M K0 (13) = 1 1 + 2×70.0 kg_{120 kg} × 1080 J = 498 J (14) The kinetic energy decreases due to the negative work done by friction (between the parachutist and turntable) on the parachutist and the turntable.

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13-3: Gyroscope

Young and Freedman (13th edition) exercise 10.55, page 337.

(same as Young and Freedman (12th edition) exercise 10.51, page 348.)

A gyroscope is precessing about a vertical axis. Describe what happens to the precession angular speed if the following changes in the variables are made, with all other variables remaining the same:

(a) the angular speed of the spinning flywheel is doubled; Solution: Apply equation 10.33, Ω = dφ dt = τz Lz = W r Iω (15)

where W is the weight.

ω → 2ω ⇒ Ω → 1

2Ω (16)

(b) the total weight is doubled; Solution:

W → 2W ⇒ Ω → 2Ω (17)

(Here we assume that the added weight does not change I or r) (c) the moment of inertia about the axis of the spinning flywheel is doubled;

Solution:

I → 2I ⇒ Ω → 1

2Ω (18)

(assuming this does not change W or r)

(d) the distance from the pivot to the center of gravity is doubled. Solution:

r → 2r ⇒ Ω → 2Ω (19)

(e) What happens if all four of the variables in parts (a) through (d) are doubled? Solution:

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13-4: Inelastic collision

A slender uniform metal rod with mass M and length ` is pivoted at its upper end and hangs vertically. A small blob with mass m traveling horizontally with velocity v0 strikes

the lower end of the rod and sticks to it.

(a) Find the angular momentum of the blob with respect to the pivot prior to the colli-sion.

Solution:

Angular momentum of the blob with respect to the pivot p, ~

Lblob = mv0`ˆk (20)

(b) Find the angular momentum of the blob+rod system with respect to the pivot im-mediately after the collision.

Solution:

Using angular momentum conservation immediately after the collision. The an-gular momentum of the blob + rod, ~L satisfy

~

L = 0 + ~Lblob= mv0`ˆk (21)

(c) Calculate the moment of inertia of the blob+rod system around the pivot immedi-ately after the collision.

Solution:

About the pivot, net net moment of inertia is I = m`2+1 3M ` 2 ₌ m + 1 3M `2 (22)

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Solution:

From ~L = I ~ω , the angular velocity of the blob + rod system right after the collision is ~ ω = L I ˆ k = m m + M₃ v0 ` ˆ k (23)

(e) Take θmax to be the maximum angle to the vertical made by the blob+rod system

following the collision. What is difference in the gravitational potential energy of the blob+rod system between θmax and θ = 0?

Solution:

The difference in gravitational potential energy is

∆U = U (θ = θmax) − U (θ = 0) (24)

= M g`

2(1 − cos θmax) + mg`(1 − cos θmax) (25) = 1

2M + m

g`(1 − cos θmax) (26)

(f) Find θmax in terms of the given quantities: M, `, m, v0 and (as always) g.

Solution:

Use energy conservation, from right after collision to the maximum angle θmax,

∆U = −∆K (27) 1 2M + m g`(1 − cos θmax) = 1 2Iω 2 ₌ L2 2I = (mv0`)2 2 m + M₃ `2 (28) 1 − cos θmax = m2 2 m + M₃ m +M₂ v₀2 g` (29) θmax = arccos 1 − m2 2 m + M₃ m + M₂ v2₀ g` ! (30) 13-5: Stopping a bicycle

An idealized bicycle+rider consists of a block representing the rider, and two wheels of radius R with centers separated by distance L. Suppose that the center of mass of the system (total mass M ) is midway between the wheels at a height 2R above the ground. The rider squeezes the front brakes, locking the front wheel. What is the maximum deceleration such that the system slows down without the rear wheel leaving the ground (and the rider going over the handlebars)?

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Solution:

When the rider squeezes the front brake, the front wheel is locked and starts slipping. The bike decelerates with a magnitude of a due to the friction between the front wheel and the ground. From Newton’s 2nd Law, f = M a. Since f = µNf, as a increases,

Nf increases (the weight M g “shifts” to the front.

Apart from the deceleration a, friction f also contributes to a torque about the c.m. that tends to lift the rear wheel; this torque is cancelled by the torque from Nf and

Nr when a is small. But when a is big enough, Nf → M g and Nb → 0. This is when

the rear wheel leaves the ground.

In addition, when the rear wheel is just about to leave the ground, we have ay = 0,

α = 0 (about c.m.), so    Nb = 0 Nb + Nf − M g = 0 P τ = NfL₂ − g(2R) = 0 =⇒ f = M g L 4R (31) so a = f M = g L 4R (32)

This is the maximum deceleration such that the bike slows down without the rear wheel leaving the ground (bike tilting around the front wheel).

13-6: Backspin on cue ball

In a game of billiards, a skillful player hits a cue ball with just enough backspin so that the ball stops after traveling a short distance away from where is was hit. Suppose the initial velocity of the ball was v0 (after being hit). Assume the cue ball is a solid sphere

with mass m and radius r. How much backspin (that is, what negative initial angular velocity ω0) would be required for this to occur? Solve this problem two ways:

(a) Consider the force of friction and use equations describing the linear kinematics and rotational kinematics.

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Solution:

Newton’s 2nd Law:

f = µkN = µkmg =⇒ ax = −

f

m = −µkg (33)

Torque about the c.m.:

τ = −f r = −µkmgr (34) ⇒ α = τ I = −µkmgr (2/5)mr2 (35) = −5µkg 2r (36) Kinematics: v = v0+ axt, ω = ω0+ αt (37) Thus when v = 0, t = v0

µkg. For the ball to stop, ω also has to vanish at the same

time. That is, 0 = ω0+ αt.

ω0 = −αt = 5µkg 2r · v0 µkg = 5v0 2r (38)

(b) Find a reference point about which the net torque is zero and use the concept of conservation of angular momentum. In this approach you will probably want to use the theorem that says when a rigid body translates and simultaneously rotates about an axis of symmetry, then the angular momentum about any fixed point S can be written as

~

L = ~rcm× ~P tot+ Icm~ω ,

where rcm is a vector from the reference point S to the center of mass, ~P tot is the

total momentum of the object, Icm is the moment of inertia of the object about the

rotation axis through the center of mass, and ~ω is the angular velocity.

You (obviously) should get the same answer in both methods. Which method was easier?

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Choose the reference point to be Q along the direction of ~v0 on the table. Notice

that around Q, f contributes 0 torque whereas the contribution from N and mg cancel, therefore the net torque around Q is 0. Angular momentum about this point is conserved. The initial angular momentum is

~ L1 = Icm~ω0 | {z } (i) + m~r × ~v0 | {z } (ii) = Icmω0k + (−mvˆ 0r)ˆk (39)

where (i) is the angular momentum about the c.m. and (ii) is the angular mo-mentum due to the motion of the center of mass.

The final angular momentum ~L2 = 0, since the motion stops completely. Since

angular momentum about Q is conserved, we have ~L1 = ~L2 and therefore ~L1 =

0, which implies ω0 = mv0r Icm = mv₂ 0r 5mr 2 = 5v0 2r (40) 13-7: Geosynchronous satellites

Young and Freedman (13th edition) problem 13.51, page 432.

(same as Young and Freedman (12th edition) problem 12.55, page 415.)

Many satellites are moving in a circle in the earth’s equatorial plane. They are at such a height above the earth’s surface that they always remain above the same point. (a) Find the altitude of these satellites above the earth’s surface. (Such an orbit is said

to be geosynchronous.) Solution:

For satellites to remain above the same point on the earth means they have to have the same orbital period as earth

T = 1 d = (24 × 3600) s = 8.64 × 104s (41) Now using equation 13.12

T = 2πr 3/2 √ GmE =⇒ r = T 2_Gm E (2π)2 13 (42) is the radius of the satellite’s orbit.

r = (8.64 × 10 4_s)2_{(6.673 × 10}−11_{N m}2_kg−2_{) (5.97 × 10}24_kg) 4π2 !1₃ (43) = 4.22 × 107m (44)

The altitude of the satellites h satisfy r = h + RE, thus

h = r − RE = 4.22 × 107m − 6.38 × 106m = 3.58 × 107m (45)

(b) Explain, with a sketch, why the radio signals from these satellites cannot directly reach receivers on earth that are north of 81.3◦ N latitude.

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Solution:

Draw the tangent line to the earth from the same satellite as in the picture. Radio signals from the satellite are blocked by the earth from reaching higher latitudes beyond the tangent point.

θ = cos−1 RE r = cos−1 6.38 × 10 6_m 4.22 × 107_m = 1.42 rad (46) = 81.4◦ (47)

13-8: Gravity near surface of the Earth

There are two equations from which a change in the gravitational potential energy U of the system of a mass m and the earth can be calculated. One is U = mgy (Eq. 7.2 of Y&F 13th edition). The other is U = −GmEm/r (Eq. 13.9). As shown in Section 13.3,

the first equation is correct only if the gravitational force is a constant over the change in height ∆y. The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in U between a mass at the earth’s surface and a distance h above it using both equations, and find the value of h for which Eq. (7.2) is in error by 1%. Express this value of h as a fraction of the earth’s radius, and also obtain a numerical value for it. [In the 12th edition, the references are Eq. 7.2, Eq. 12.9, and Section 12.3.]

Solution:

The fraction error is expressed as

α = 1 − mgh GmEm RE − GmEm RE+h (48) = 1 − gh GmE 1 1 RE − 1 RE+h = 1 − g GmE RE(RE+ h) (49) From Equation 13.4, g = GmE R2 E , so α = 1 −RE+ h RE = − h RE (50)

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Therefore if the fraction error is 1%, |α| = 1% h

RE

= 0.01 =⇒ h = 0.01RE = 6.4 × 104m (51)

13-9: Falling hammer

A hammer with mass m is dropped from rest from a height h above the earth’s surface. This height is not necessarily small compared with the radius of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, RE, and mE, the mass of the

earth. Solution:

Apply conservation of energy, and note that we need to use (13.9) U = −GmEm

r for

gravitational potential energy since h is not necessarily small compared to RE

K1+ U1 = K2+ U2, U1 = − GmEm h + RE , U2 = − GmEm RE (52) ⇒ 0 + −GmEm h + RE = 1 2mv 2 + −GmEm RE (53) ⇒ v2 = 2Gme 1 RE − 1 RE+ h (54) v = s 2GmEh RE(RE + h) (55) 13-10: Spacecraft escape

(same as Young and Freedman (12th edition) problem 12.69, page 416.) The astronomical data needed for this problem can be found in Appendix F of the textbook, in either the 12th or 13th editions.

A spacecraft is to be launched from the surface of the earth so that it will escape from the solar system alto-gether.

(a) Find the speed relative to the center of the earth with which the spacecraft must be launched. Take into consideration the gravitational effects of both the earth and the sun, and include the effects of the earth’s orbital speed, but ignore air resistance.

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Solution:

Apply energy conservation to find the escape velocity from the solar system, assume that the gravitational potential energy between the spacecraft and the earth/the sun dominate. RE is the radius of the earth; RES is the distance

between the earth and the sun. 1 2mv 2₋GmmE RE −GmmS RS = 0 (56)

The escape velocity v (in the reference frame of a distance star) v = s 2G mE rE + mS RS (57) = s 2 × 6.67 × 10−11N m2kg−2× 5.97 × 10 24_kg 6.38 × 106_m + 1.99 × 1030kg 1.50 × 1011_m (58) = 4.35 × 104m/s (59)

Now assume the spacecraft is launched in the direction of the earth orbital motion around the sun, the speed relative to the center of the earth is

v1 = v − 2πRES T = 4.35 × 10 4_{m/s −} 2π × 1.50 × 1011m (365 × 24 × 3600) s (60) = 1.36 × 104m/s (61)

(b) The rotation of the earth can help this spacecraft achieve escape speed. Find the speed that the spacecraft must have relative to the earth’s surface if the spacecraft is launched from Florida at the point shown in the figure. The rotation and orbital motions of the earth are in the same direction. The launch facilities in Florida are 28.5◦ north of the equator.

Solution:

The rotation and orbital motion of the earth are in the same direction. The speed of a point on the surface of the earth at an angle θ from the equator is v = 2πREcos φ

TE , where TE = (24 × 3600) s = 8.64 × 10

4_{s. Therefore launched from}

Florida, the spacecraft must have a speed v2 relative to the earth’s surface.

v2 = v1− 2πRecos 28.5◦ TE (62) = 1.36 × 104m/s − 2π(6.38 × 10 6_{m) cos 28.5}◦ 8.64 × 104s (63) = 1.32 × 104m/s (64)

(c) The European Space Agency (ESA) uses launch facilities in French Guiana (immedi-ately north of Brazil), 5.15◦ north of the equator. What speed relative to the earth’s surface would a spacecraft need to escape the solar system if launched from French Guiana?

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Solution:

Similarly, launched from French Guiana, the space craft must have an escape speed of v3 relative to the earth’s surface,

v3 = v1− 2πRecos 5.15◦ TE (65) = 1.36 × 104m/s − 2π(6.38 × 10 6_{m) cos 5.15}◦ 8.64 × 104s (66) = 1.31 × 104m/s (67)

Super-Challenge Problem 1: Elliptical orbit

Adapted from Young and Freedman (13th edition) problem 13.77, page 434. (same as Young and Freedman (12th edition) problem 12.77, page 417.)

For up to 10 points of extra credit, you may do the following totally optional problem. For a description of how extra credit problems will be processed, see Problem Set 9.

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is at a distance rp from the earth’s center; at the high point, or apogee, it

is a distance ra from the earth’s center.

(a) What is the period of the spacecraft’s orbit? Solution: Since 2a = ra+ rp, we have T = 2πa 3/2 √ GmE = 2π GmE r_a+ rp 2 3₂ (68)

(b) Using conservation of angular momentum, find the ratio of the spacecraft’s speed at perigee to its speed at apogee.

Solution:

Using conservation of angular momentum about the earth,

mrava = mrpvp (69) vp va = ra rp (70)

(c) Using conservation of energy, find the speed at perigee and the speed at apogee. Solution:

Apply conservation of energy to apogee and perigee

Ka+ Ua = Kp+ Up (71) 1 2mv 2 a− GmEm ra = 1 2mv 2 p − GmEm rp (72)

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v_p2− v2 a = 2GmE 1 rp − 1 ra = 2GmE(ra− rp) rarp (73)

use result from (b)

v_p2 1 − rp ra 2! = 2GmE(ra− rp) rarp (74) v_p2 = 2GmE ra− rp rarp r2 a r2 a− rp2 = 2GmEra rp(ra+ rp) (75) vp = s 2GmEra rp(ra+ rp) (76) va = rpvp ra = s 2GmErp ra(ra+ rp) (77)

(d) It is necessary to have the spacecraft escape from the earth completely. If the space-craft’s rockets are fired briefly at perigee, producing a sudden change in the rocket’s speed, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use? Solution:

For the spacecraft to escape, E has to be at least 0, K + U = 0. At apogee: 1 2m˜v 2 p − GmEm rp = 0 =⇒ ˜vp = s 2GmE rp (78) the speed has to be increased by

∆vp = ˜vp− vp = s 2GmE rp 1 − r _r a ra+ rp (79) At perigee: 1 2m˜v 2 a− GmEm ra = 0 =⇒ v˜a= r 2GmE ra (80) the speed has to be increased by

∆va = ˜va− va= r 2GmE ra 1 − r _r p ra+ rp (81)

It is not easy to see which of these two expressions is larger, but if one inserts numbers one finds immediately that ∆v is always smaller at perigee. This implies

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that less fuel would be needed to fire the rockets at perigree, since the fuel needs are determined entirely by the change in velocity. (Note that the spacecraft flies essentialy through a vacuum, so it does not interact with the Earth, and its speed relative to the Earth is of no relevance. The same change in velocity will always require the same amount of fuel, regardless of the initial velocity relative to the Earth.)

Super-Challenge Problem 2: Navigation from Earth to Mars Young and Freedman (13th edition) problem 13.87, page 435.

For up to 10 points of extra credit, you may do the following totally optional problem. For a description of how extra credit problems will be processed, see Problem Set 9.

Figure P13.87

The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit (Fig. P13.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the departure planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet.

(a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth?

Solution:

The orbital radius of the Earth is rE = 1.50 × 1011m, and the orbital radius of

Mars is rM = 2.28 × 1011m. One can see from the figure that the major axis of

the Hohmann transfer orbit is just the sum of these two, so the semi-major axis for the elliptical Hohmann transfer orbit is given by

a = rM + rE

2 = 1.89 × 10

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If the spacecraft does not fire its engines, it will of course remain in its circular orbit at radius rE. To reach the larger radii of the Hohmann transfer orbit, it

must increase its speed (and energy), and so the rockets must be fired in the opposite direction of the motion. Once the spacecraft reaches the orbit of Mars, if it did not fire its rockets, it would continue on the elliptical orbit and fall back to smaller radii, as it approached once more the radius of the Earth’s orbit. To maintain the larger radius of Mars’ orbit, it must again increase its speed (and energy), which means that the rockets must again be fired in the direction opposite that of the motion.

For the returning flight from Mars to Earth, the precedure is reversed, and the rockets must be fired in the direction of the motion so as to decreased the speed and thus decreased the mechanical energy.

(b) How long does a oneway trip from the the earth to Mars take, between the firings of the rockets?

Solution:

The time of the oneway trip would be one half the period of the transfer orbit. T = 2πa 3/2 √ Gms (83) t = T 2 = πa3/2 √ Gms = π (1.89 × 10 11_m)32 p 6.67 × 10−11N m2kg−2× 1.99 × 1030_kg (84) = 2.24 × 107s = 259 d (85)

(c) To reach Mars from the earth, the launch must be timed so that Mars will be at the right spot when the spacecraft reaches Mars’s orbit around the sun. At launch, what must the angle between a sun–Mars line and a sun–earth line be? Use data from Appendix F.

Solution:

During this time t, Mars will pass through an angle of θ = 2π t

TM

= 2π ×259 d

687 d (86)

= 2.37 rad = 136◦ (87)

Since the spacecraft passes through an angle of 180◦about the Sun, at launch, the angle between the Sun-Mars line and the Sun-Earth line must be 180◦− 136◦ = 44◦. The earth must be 44◦ behind Mars in the direction of the orbital motion.